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For a 3x3 matrix $C$, it is given that

$$C^3+I=3C^2-C$$

I am then required to prove that $C$ is invertible.

I have attempted a proof, below, but I am not sure it is valid or if there is a better solution.

Attempted proof

$$C^3 + I = 3C^2 - C$$ $$I = - C^3 +3C^2-C$$

If it is assumed that $C^{-1}$ exists then

$$I = C^{-1}(-C^4+3C^3-C^2)$$ If $C^{-1}$ is defined, then $I=C^{-1}C$; therefore test whether

$$C \stackrel{!}{=} -C^4 + 3C^3 - C^2$$ $$ 0 = -C^4 + 3C^3 - C^2 - C$$ $$ C = 0, 1, 1\pm\sqrt{2}$$

Is this at all in the right direction?

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6  
$C$ is a matrix, how can it be $0,1,1\pm \sqrt{2}$? –  vadim123 Jun 13 at 22:46

6 Answers 6

up vote 17 down vote accepted

Note that $$ I=-C^3+3C^2-C=C(-C^2+3C-I) $$ Hence $C$ is invertible and $C^{-1}=-C^2+3C-I$.

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Subtract to get $$I=-C^3+3C^2-C$$ Then factor to get $$I=C(-C^2+3C-I)$$ Now you have $I=CD$, for $D=-C^2+3C-I$. Hence $C$ is invertible.

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$$C^3+I=3C^2-C$$ $$I=-C^3-3C^2-C$$ $$\det(I)=\det(-C^3+3C^2-C)=\det(C)\det(-C^2+3C-I)$$ Assume $C$ is not invertible, i.e. $\det(C)=0$, then from the previous equation you would obtain $\det(I)=0$, that is a contradiction, since $\det(I)=1$.

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Quick clean and efficient; I love this. –  Geoff Naylor Jun 14 at 0:57
1  
Why prove by contradiction? –  Tim Seguine Jun 14 at 10:36
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@GeoffNaylor That's a very roundabout way of proving invertibility. Brian's answer is more direct. –  Jubobs Jun 14 at 15:23

If $C$ is not invertible, it has a non-trivial null space and hence a non-trivial eigenvector $v$ with eigenvalue $\lambda=0$.

We have $(C^3+I)v=(3C^2-C)v$ whence $(\lambda^3+1)v=(3\lambda^2-\lambda)v$ and setting $\lambda=0$ we see that $v=0$, which is a contradiction.

Easiest (and in fact more general) to use the explicit inverse in the other answers, but this is a way to use the eigenvalues as suggested in your answer.

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The original assumption implies $C$ annihilates the polynomial $X^3-3X^2+X+1$.

But $X^3-3X^2+X+1=(X-1)(X-\sqrt2-1)(X+\sqrt2-1)$

Therefore, $C$ is diagonalizable, but most importantly its determinant is the product of its eigenvalues, hence $|\det(C)|=1$ and $C$ is invertible.


For the sake of generality, for $n\times n$ matrix, $C$ still annihilates $(X-1)(X-\sqrt2-1)(X+\sqrt2-1)$.

By Primary Decomposition Lemma , $C$ is diagonalizable with eigenvalues $1$, $1-\sqrt{2}$, $1+\sqrt{2}$

Hence it has non-zero determinant.

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$C$ is a 3x3 matrix, so it's characteristic polynomial is $ax^3 + bx^2 + cx + d$.

A matrix is invertible iff it doesn't have a zero eigenvalue, which occurs iff $d \ne 0$.

Every matrix satisfies it's characteristic polynomial.

Your given polynomial is a 3rd degree polynomial with $d \ne 0$, thus the characteristic polynomial is as well, thus $C$ doesn't have an eigenvalue of $0$, thus it is invertible.

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